Home  Work  Solutions            PHYS111.02  SFSU  Eradat                                    [WH6  CHAPTERS  16-­‐18]   1    +WHW6: Chapter 16: 11,19,42,55, Chapter 17: 17,28,72, Chapter 18: 27,38,51  Chapter  16  11.     Picture  the  Problem:  Our  sketch  shows  the  pressure  

in  the  gas  thermometer  as  a  function  of  the  temperature.    Note  that    at  T  =  100°C  the  pressure  is  227  mmHg  and  the  pressure  extrapolates  to  zero  at  the  temperature  −273.15°C.    Assuming  ideal  behavior  we  want  to  find  the  temperature  associated  with  a  pressure  of  162  mmHg.    

 

  Strategy:    We  assume  that  the  pressure  lies  on  a  straight  line.  Using  the  known  pressures  at  100°C  and  −273.15°C,  calculate  the  rate  at  which  pressure  increases  as  a  function  of  temperature.    Calculate  the  temperature  at  162  mmHg  using  this  rate.  

  Solution:  1.    Divide  the  change  in  pressure  by  the  change  in  temperature  to  obtain  the  rate:  

rate = 227 mmHg − 0100°C − −273.15°C( ) = 0.60833 mmHg/C°  

 2.    Solve  the  rate  equation  for  the  temperature:  

rate =P − P0

T − T0

⇒ T = T0 +P − P0

rate

 

 3.    Insert  given  data:    

T = 100°C + 162 mmHg − 227 mmHg

0.60833 mmHg/C°= − 6.85°C  

  Insight:  The  gas  thermometer  is  first  placed  in  the  boiling  water  to  calibrate  it,  because  water  boils  at  a  known  temperature.    Once  the  thermometer  is  calibrated,  the  pressure  variation  can  accurately  give  temperatures  over  a  wide  range.    

 19.     Picture  the  Problem:  A  steel  bar  has  a  larger  diameter  than  that  

of  an  aluminum  ring  that  you  would  like  to  slip  over  the  bar.         Strategy: Use equation 14-6 to calculate the temperature at which the

ring’s inner diameter will equal the diameter of the bar. The coefficient of thermal expansion is given in Table 16-1.

  Solution:  1.  (a)    The  ring  should  be  heated.  Imagine  that  the  ring  is  cut  and  “unrolled.”  It  would  be  a  rectangle.  If  the  rectangle  is  heated,  it  will  expand  along  its  length  and  width.  Its  length  is  the  circumference  of  the  ring.  Since  the  length  of  the  rectangle  increases,  the  circumference  of  the  circle  increases,  and  therefore,  so  does  its  diameter.  

 

  2.  (b)  Solve  equation  16-­‐4  for    the  change  in  temperature:  

Δd = α dΔT ⇒ ΔT = T − T0 =Δdαd

 

 3.  Solve  for  the  final  temperature:  

T = T0 +Δdαd0

= 10.00 °C + 4.040 cm – 4.000 cm

2.4 ×10−5 C°( )−1( ) 4.000 cm( )= 430°C

    Insight:  When  the  aluminum  is  heated  to  430°C  it  will  slip  over  the  steel  rod.    As  it  cools  back  down  it  will  

shrink  to  form  a  tight  bond  with  the  steel.      42.     Picture  the  Problem:  Heat  transfers  from  a  hot  horseshoe  to  the  cold  water.    This  decreases  the  

temperature  of  the  horseshoe  and  increases  the  temperature  of  the  water  until  the  water  and  horseshoe  are  at  the  same  equilibrium  temperature.  

  2  

  Strategy:  Since  only  two  objects  are  transferring  heat,  use  equation  16-­‐15  to  calculate  the  equilibrium  temperature.    To  determine  which  object  would  cause  a  larger  final  temperature,  you  should  compare  the  heat  capacities  heats  of  the  two  objects.    The  object  with  the  higher  heat  capacity  will  have  more  heat  to  transfer  to  the  water,  causing  the  final  temperature  to  be  greater.  

  Solution:  1.  (a)  Insert  given  data  into  equation  16-­‐15:  

T =mhchTh + mwcwTw

mhch + mwcw

=0.50 kg 448 J/ kg ⋅K( )⎡⎣ ⎤⎦ 450 °C + 25 kg 4186 J/ kg ⋅K( )⎡⎣ ⎤⎦ 23 °C

0.50 kg( ) 448 J/ kg ⋅K( )⎡⎣ ⎤⎦ + 25 kg( ) 4186 J/ kg ⋅K( )⎡⎣ ⎤⎦T = 24 °C

 

  2.  (b)  Write  the  heat  capacities  of  the  lead  and  iron:  

CPb = 1 kg ×128J/ kg ⋅K( )⎡⎣ ⎤⎦ = 128 J/K

CFe = 0.50 kg × 448 J/ kg ⋅K( )⎡⎣ ⎤⎦ = 224 J/K  

  3.  Compare  the  heat  capacities:   Since  the  heat  capacity  of  the  lead  is  less  than  the  heat  capacity  of  the  iron,  the  final  temperature  will  be  less  than  24°C.  

  Insight:  Even  though  the  lead  had  twice  the  mass  of  the  iron,  the  specific  heat  of  lead  is  small  enough  that  the  heat  capacity  of  the  iron  was  larger  than  the  heat  capacity  of  the  lead.  

 55.     Picture  the  Problem:  Two  metal  bars,  of  equal  lengths,  are  

connected  in  parallel.    Heat  transfers  across  both  bars.    The  diameter  of  the  lead  rod  is  known.    We  want  to  know  what  diameter  of  the  copper  rod  is  necessary  for  the  rate  of  heat  transfer  through  both  rods  to  be  33.2  J/s.  

  Strategy:  Set  the  total  rate  of  heat  transfer  equal  to  the  sum  of  the  heat  transfers  through  each  rod.    Using  equation  16-­‐16,  solve  for  the  diameter  of  the  copper  rod.    

  Solution:  1.  Sum  the  heat    transfer  rates  of  the  two    rods,  using  equation  16-­‐16:  

Qtotal

t=

QCu

t+

QPb

t= kCu ACu

ΔTL

⎛⎝⎜

⎞⎠⎟+ kPb APb

ΔTL

⎛⎝⎜

⎞⎠⎟

= kCu

π4

dCu2⎛

⎝⎜⎞⎠⎟+ kPb

π4

dPb2⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

ΔTL

⎛⎝⎜

⎞⎠⎟= kCudCu

2 + kPbdPb2⎡⎣ ⎤⎦

πΔT4L

⎛⎝⎜

⎞⎠⎟

 

 

2.  Solve  for  the  diameter    of  the  copper  rod:  

dCu =

Qtotal

t⎛

⎝⎜⎞

⎠⎟4LπΔT

⎛⎝⎜

⎞⎠⎟− kPbdPb

2⎡

⎣⎢⎢

⎤

⎦⎥⎥

kCu

=

33.2 J/s( ) 4 0.650 m( )π 112 − 21( ) °C − 34.3W/ m ⋅K( )⎡⎣ ⎤⎦ 0.0276 m( )2

395W/ m ⋅K( ) = 2.64 cm

    Insight:  The  diameters  of  both  rods  are  about  the  same.    However,  because  the  thermal  conductivity  of  

the  copper  is  over  10  times  the  thermal  conductivity  of  lead,  over  90%  of  the  heat  passes  through  the  copper.  

       

Home  Work  Solutions            PHYS111.02  SFSU  Eradat                                    [WH6  CHAPTERS  16-­‐18]   3    Chapter  17  17.     Picture  the  Problem:  The  Martian  atmosphere  can  be  considered  as  an  ideal  gas  with  a  given  

temperature  and  pressure.  

  Strategy:  Use  the  ideal  gas  law  (equation  17-­‐2)  to  solve  for  the  number  of  molecules  per  volume.    Compare  the  results  for  the  Martian  atmosphere  and  Earth’s  atmosphere  at  standard  temperature  and  pressure.  

 Solution:  1.  (a)  Convert  the    Martian  temperature  to  Kelvin:  

Tc =59

Tf − 32( ) = 59

− 64 − 32( ) = −53.3°C

Tk = 273.15− 53.3 K( ) = 219.8 K

 

  2.  Solve  the  ideal  gas  law  for    number  of  molecules  per  volume:  

NV

= PkT

= 0.92 ×103 Pa1.38 ×10−23J/K( ) 219.8 K( ) = 3.0 ×1023 molecules m3  

  3. (b) The number of molecules per volume on Earth is greater than that on Mars. The temperatures of Earth and Mars have the same order of magnitude, but the pressure on Earth is far greater than that on Mars.

  4.  (c)  Solve  for  the  number  of    molecules  per  volume  on  Earth:  

NV

= PkT

= 1.01×105 Pa1.38 ×10–23J/K( ) 273.15 K( ) = 2.68 ×1025 molecules m3

 

  Insight:  The  density  of  molecules  on  Earth  is  about  100  times  the  density  of  molecules  on  Mars.    When  landing  a  rover  on  Mars,  the  scarce  atmosphere  results  in  a  higher  terminal  velocity,  even  with  a  parachute.  

 28.     Picture  the  Problem:  An  ideal  gas  is  confined  in  a  chamber.    The  gas  expands  to  twice  the  original  

volume  at  constant  temperature.     Strategy:  Use  the  ideal  gas  law,  equation  17-­‐5,  to  determine  the  pressure  and  equation  17-­‐12  to  calculate  

the  average  kinetic  energy.      

Solution:  1.  (a)  Solve  equation  17-­‐5  for  P:  

P = nRTV

=3 mol 8.31J/ mol ⋅K( )⎡⎣ ⎤⎦ 273.15+ 295 K( )

0.0035 m3 = 4.0 ×106 Pa

 

  2.  (b)  Insert  the  given  data  into  equation    17-­‐12  for  the  kinetic  energy:  

Kav = 3

2 kT = 32 1.38 ×10–23J/K( ) 273.15+ 295 K( ) = 1.18 ×10–20 J

 

  3.  (c)    Pressure  decreases  by  a  factor  of  2  because  pressure  is  inversely  proportional  to  volume.  The  average  kinetic  energy  of  a  molecule  remains  the  same,  because  it  depends  only  on  temperature.  

  Insight:  The  average  kinetic  energy  of  the  gas  molecules  depends  on  the  temperature,  not  on  the  pressure,  volume,  or  number  of  moles  present  in  the  gas.    As  the  gas  expands,  the  number  of  collisions  with  the  walls  decreases,  causing  a  decrease  in  pressure.      

   72.     Picture  the  Problem:  Ice  is  dropped  into  some  water.    Heat  transfers  from  the  water  and  melts  the  ice  

until  the  water  and  ice  are  in  thermal  equilibrium.     Strategy: Assume that the equilibrium temperature is 0°C and that not all of the ice has melted. Using this

assumption and equation 16-13, calculate the heat lost by the water as it cools to 0°C. Set that heat equal to the latent heat of fusion of the ice, equation 17-20, to calculate the mass of ice that melts. If that mass is less than the total mass of ice, then our assumption is correct. To calculate the minimum temperature for which all of the ice will melt, set the latent heat of the melting ice equal to the heat lost by the water and solve for the initial

  4  

temperature.   Solution:  1.  (a)  Calculate  the  heat  lost  by  the  

water:   Q = mcΔT = 0.33 kg 4186 J/ kg ⋅K( )⎡⎣ ⎤⎦ 14C°( ) = 1.93×104 J

   

2.  Calculate  the  mass  of  the  melted  ice:       m = Q

Lf

= 1.93×104 J33.5×104 J/kg

= 0.058 kg

  3.  Calculate  the  remaining  ice:   mf = 0.075 kg − 0.058 kg = 0.017 kg

  4.  (b)  Set  the  latent  heat  required  to  melt  the  ice  equal  to  the  heat  lost  by  the  water:      

mwc T − 0°C( ) = mice L  

 5.  Solve  for  the  initial  temperature  of  the  water:  

T =mice Lmwc

=0.075 kg( ) 33.5×104 J/kg( )0.33 kg( ) 4186 J/ kg ⋅K( )⎡⎣ ⎤⎦

= 18°C  

  Insight:  A  small  amount  of  ice  can  cool  a  large  drink  because  the  latent  heat  of  fusion  of  water  is  quite  large.  

 Chapter  18  27.     Picture  the  Problem:  An  ideal  gas  completes  a  three-­‐process  

cycle.  

  Strategy:  Use  the  ideal  gas  law  to  calculate  the  temperature  at  each  state.    Solve  the  first  law  of  thermodynamics  for  the  heat  absorbed  in  each  process.  

  Solution: 1. (a) Solve the ideal gas law for temperature:

T = PVnR

 2.  Solve  for TA :  

TA =150 ×103 Pa( ) 1.00 m3( )

67.5 mol( ) 8.31 J mol ⋅K( )= 267 K

   

 3.  Solve  for TB :  

TB =

50 ×103 Pa( ) 4.00 m3( )67.5 mol( ) 8.31 J mol ⋅K( ) = 357 K  

 4.  Solve  for TC :  

TC =

50 ×103 Pa( ) 1.00 m3( )67.5 mol( ) 8.31 J mol ⋅K( ) = 89.1 K  

  5.  (b)   A → B :The  temperature  rises  and  the  gas  does  work,  so  heat  enters  the  system.       B→ C :  The  temperature  drops  and  work  is  done  on  the  gas,  so  heat  leaves  the  system.       C → A :  The  temperature  rises  and  no  work  is  done  on  or  by  the  gas,  so  heat  enters  the  system.  

  6.  (c)  Solve  the  first  law  for  Q:   Q = ΔU +W = 32 nRΔT +W

  7.    Insert  values  for   A → B :  

Q = 32 67.5 mol( ) 8.31 J mol ⋅K( ) 357 K − 267 K( )+ 1

2 150 kPa + 50 kPa( ) 3.00 m3( ) = 376 kJ

  8.    Insert  values  for   B→ C :  

Q = 32 67.5 mol( ) 8.31 J mol ⋅K( ) 89.1 K − 357 K( )+ 50 kPa −3.00 m3( ) = −375 kJ

Home  Work  Solutions            PHYS111.02  SFSU  Eradat                                    [WH6  CHAPTERS  16-­‐18]   5       9.    Insert  values  for   C → A :  

Q = 3

2 67.5 mol( ) 8.31 J mol ⋅K( ) 267 K − 89.1 K( ) + 0 = 150 kJ

  Insight:  The  net  work  done  during  this  cycle  is  equal  to  the  net  heat  absorbed,  or  W  =  150  kJ.    Although  the  sum  of  the  heats  in  part  (d)  appears  to  be  151  kJ,  that  is  an  artifact  of  significant  digits  and  rounding  issues,  it’s  exactly  150  kJ.  

   38.     Picture  the  Problem:  A  monatomic  ideal  gas  is  expanded  at  

constant  pressure  by  a  fixed  change  in  volume.    This  process  is  done  twice  with  differing  initial  volumes.  

Strategy:    Use  the  area  under  the  PV  plot  to  calculate  the  work  done  in  each  process.    

 

Solution:  1.  (a)  Write  the  work    as  area  under  the  PV  plot:   W = PΔV  

   

2.    Solve  numerically:   W = 160 ×103 Pa( ) 8600 cm3 − 5400 cm3( ) 1×10−6 m3

cm3

⎛

⎝⎜⎞

⎠⎟= 0.51 kJ

 

  3.  (b)  Work  is  directly  proportional  to  the  change  in  volume.  Therefore,  the  work  done  by  the  gas  in  the  second  expansion  is  equal  to  that  done  in  the  first  expansion.  

 4.  (c)  Solve  numerically:  

W = 160 ×103 Pa( ) 5400 cm3 − 2200 cm3( ) 1×10−6 m3

cm3

⎛

⎝⎜⎞

⎠⎟= 0.51 kJ

    Insight:  Because  the  work  is  proportional  to  the  change  in  volume,  increasing  the  volume  by  3200  cm3  

from  any  initial  volume  will  produce  the  same  amount  of  work.    51.     Picture  the  Problem:  The  efficiency  of  a  Carnot  engine  is  increased  by  lowering  the  cold  temperature  

reservoir.     Strategy:  Use  equation  18-­‐13  to  solve  for  the  temperature  of  the  cold  temperature  reservoir  using  the  

initial  efficiency,  and  then  the  new  efficiency.     Solution:  1.  (a)  Solve  equation  18-­‐13  for Tc :    

Tc = Th 1− emax( ) = 545 K 1− 0.300( ) = 382 K  

  2.  (b)  The  efficiency  of  a  heat  engine  increases  as  the  difference  in  temperature  of  the  hot  and  cold  reservoirs  increases.  Therefore,  the  temperature  of  the  low  temperature  reservoir  must  be  decreased.  

  3.  (c)  Insert  the  new  efficiency:     Tc = 545 K 1− 0.400( ) = 327 K  

  Insight:  The  efficiency  of  a  Carnot  engine  is  increased  when  the  temperature  difference  between  the  two  reservoirs  increases.    When  the  hot  temperature  reservoir  is  fixed,  the  efficiency  can  be  increased  by  lowering  the  temperature  of  the  cold  reservoir.